# Conjugacy classes for $G_1 \times G_2$

Let $$G = G_1 \times G_2$$ be the product of groups $$G_1$$ and $$G_2$$. Prove that $$|\text{conjugacy classes of G}| = |\text{number of conjugacy classes of G_1}| \cdot |\text{number of conjugacy classes of G_2}|.$$

I believe that this proof requires that the groups be finite. Otherwise, I can't say that any of them possess a finite number of conjugate classes. So, suppose each of the groups in this problem are finite. I know that the conjugacy classes have to partition the group, but their sizes can differ. Do I need to use the class equation?

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Here is a basic framework for a proof:

• Pick a conjugacy class of $$G_1$$ and a conjugacy class of $$G_2$$, and show that together they determine a single conjugacy class of $$G$$ in a very natural way
• Show that any conjugacy class in $$G$$ can be constructed this way

Consider the function $$\phi$$ that takes the conjugacy class $$[(a,b)]$$ of some element $$(a,b)\in G_1\times G_2$$ and gives you $$\phi([(a,b)])=([a],[b])$$, i.e., an ordered pair formed by the conjugacy class of $$a\in G_1$$ and the one of $$b\in G_2$$. This function (if it's well-defined) goes from the set of conjugacy classes of $$G_1\times G_2$$ to the set of ordered pairs formed by the conjugacy classes of $$G_1$$ (first component) and the conjugacy classes of $$G_2$$ (second component).

This last set has clearly cardinality $$|\text{number of conjugacy classes of G_1}| \cdot |\text{number of conjugacy classes of G_2}|$$, so all we need to do is show $$\phi$$ is a bijection. But first, we'll see it's well-defined. If $$[(a,b)]=[(c,d)]$$ then there is some $$(g,h)\in G_1\times G_2$$ such that $$(a,b)=(g,h)*(c,d)*(g,h)^{-1}=(g,h)*(c,d)*(g^{-1},h^{-1})=(g*c*g^{-1},h*d*h^{-1})$$, so $$a=g*c*g^{-1}$$ and $$b=h*d*h^{-1}$$, and thus $$[a]=[c]$$ and $$[b]=[d]$$. Therefore $$([a],[b])=([c],[d])$$ and $$\phi$$ is well defined.

Now, if $$([a_1],[b_1])=([a_2],[b_2])$$ then $$[a_1]=[a_2]$$ and $$[b_1]=[b_2]$$, so $$a_1=g*a_2*g^{-1}$$ for some $$g\in G_1$$ and $$b_1=h*b_2*h^{-1}$$ for some $$h\in G_2$$. Therefore $$(a_1,b_1)=(g,h)*(a_2,b_2)*(g,h)^{-1}$$, so $$[(a_1,b_1)]=[(a_2,b_2)]$$ and $$\phi$$ is one-to-one.

Lastly, take some $$([a],[b])$$. Considering $$[(a,b)]$$ we easily get $$\phi([(a,b)])=([a],[b])$$, so $$\phi$$ is onto.

We conclude $$|\text{conjugacy classes of G}| = |\text{number of conjugacy classes of G_1}| \cdot |\text{number of conjugacy classes of G_2}|$$.